Proof. Let X be any small subgroup of G and let A be an abelian subgroup of G having cardinality $\aleph $ . It is known (see for instance [Reference de Giovanni and Trombetti12, Lemma 3.2]) that A contains large subgroups U and V such that

$$ \begin{align*} \langle U,V \rangle = U \times V \end{align*} $$

and

$$ \begin{align*} X \cap \langle U,V \rangle = \{1\}. \end{align*} $$

As the indices $|U : U_G| $ and $|V : V_G|$ are finite, it follows that $U_G$ and $V_G$ have cardinality $\aleph $ . Therefore, the large subgroups $XU_G$ and $XV_G$ are normal-by-finite, and hence also

$$ \begin{align*} X= XU_G \cap XV_G \end{align*} $$

is finite over its core $X_G$ . Finally, if G is a generalised radical group, then all periodic homomorphic images of G are locally finite and the statement follows from [Reference Smith and Wiegold21, Theorem 2].

Proof. Clearly, $N'$ is a normal subgroup of G and the statement is obvious if $N'$ has cardinality $\aleph $ . Suppose now that $N'$ has cardinality strictly smaller than $\aleph $ . Then, $N/N'$ is a large abelian subgroup of $G/N'$ , so that $G/N'$ is a CF-group by Lemma 2.1.

Proof. Let A be a large abelian subgroup of G. Then, the core $A_G$ of A in G is a large abelian normal subgroup of G contained in the Baer radical B of G, and hence, B has cardinality $\aleph $ . However, by Lemma 2.1, G is a CF-group and so is B. Therefore, B is nilpotent and the statement follows from [Reference Smith and Wiegold21, Theorem 3].

Proof. (1) Let A be a large nilpotent subgroup of G; clearly, as every large subgroup of G is normal-by-finite, we may assume that A is normal in G. The Baer group $G/A'$ is a CF-group by Corollary 2.3 and, hence, $G/A'$ is nilpotent. Now, by a well-known nilpotency criterion of P. Hall (see [Reference Robinson19, 5.2.16]), it follows that G itself is nilpotent. Thus, there exists a positive integer s such that $\gamma _s(G)$ has cardinality $\aleph $ and $\gamma _{s+1}(G)$ has cardinality strictly smaller than $\aleph $ . Therefore, the factor group $G/ \gamma _{s+1}(G)$ contains the large abelian subgroup $\gamma _s(G) / \gamma _{s+1}(G)$ , so that $G/ \gamma _{s+1}(G)$ is a CF-group by Lemma 2.1 and, hence, it is an abelian-by-finite BCF-group (see [Reference Smith and Wiegold21, Theorem 3]).

(2) The statement follows immediately from Proposition 2.4.

Proof. Let H be any large subgroup of G; then the core $H_G$ of H in G has finite index in H and, in particular, $H_G$ is a large subgroup of G. It follows that all subgroups of the factor group $G/H_G$ are normal-by-finite. As every periodic homomorphic image of G is locally finite, the second commutator subgroup $G" H_G/ H_G$ of $G/H_G$ is locally finite (see [Reference de Giovanni and Saccomanno11, Lemma 1]).

Proof. Let $g(G" \cap Z(G)) \neq G" \cap Z(G)$ be a nontrivial element of $G"/(G" \cap Z(G))$ . Then, $g \notin Z(G)= \bigcap _{x \in G} C_G(x)$ , so there exists an element $h \in G$ such that $g \notin C_G(h).$ As the normal closure $\langle h \rangle ^ G$ is contained in $\langle h , G' \rangle $ , it is a small subgroup of G. Therefore, the index $|G : C_G(h)|$ is strictly smaller than $\aleph $ and, hence, the centraliser $C_G(h)$ is a large subgroup of G. Put $H = C_G(h)$ . Clearly, the element $g \notin H_G$ , and thus, $g(G" \cap Z(G)) \notin (G" \cap H_G)/(G" \cap Z(G))$ . Moreover, it follows from Proposition 2.6 that the factor group,

$$ \begin{align*} \frac{G"/(G" \cap Z(G))}{(G" \cap H_G)/(G" \cap Z(G))} \simeq \frac{G"}{(G" \cap H_G)}, \end{align*} $$

is locally finite and the statement is proved.

Proof. As G contains a large FC-subgroup, there exists a large subgroup K of G whose commutator subgroup $K'$ is finite (see [Reference Tomkinson23, page 155]). Therefore, $K/K'$ contains an abelian subgroup of the form $U/K' \times V/K'$ where both U and V have cardinality $\aleph $ (see [Reference De Falco, de Giovanni and Musella6, Corollary 3.2]), so that $U_G$ and $V_G$ are large subgroups of G. Moreover, the intersection $U_G \cap V_G \leq U \cap V = K'$ is finite. As all subgroups of the factor groups $G/U_G$ and $G/V_G$ are normal-by-finite, and every periodic homomorphic image of G is locally finite, it follows that the second commutator subgroups $G" U_G/ U_G$ of $G/U_G$ and $G" V_G/ V_G$ of $G/V_G$ are both locally finite (see [Reference de Giovanni and Saccomanno11, Lemma 1]). However, the factor group $G"(U_G \cap V_G) / (U_G \cap V_G)$ is isomorphic to a subgroup of the locally finite group

$$ \begin{align*} \frac{G"}{G" \cap U_G} \times \frac{G"}{G" \cap V_G} \end{align*} $$

and, hence, $G"$ is likewise locally finite.

Proof. Let X be any subgroup of N. Clearly, as N is a proper normal subgroup of G, X cannot be contranormal in G. Thus, every subgroup of N is normal-by-finite in G and the statement follows from [Reference Smith and Wiegold21, Theorem 2].

Proof. Assume first that every subgroup of G is either normal-by-finite or contranormal. If G is not of type (i), then there exists a proper subgroup B which is not normal-by-finite and, hence, it is contranormal; in particular, $G=G'B$ . Put $\overline {B}= B/(B \cap G')$ , and suppose that $\overline {B}$ is generated by two proper subgroups $\overline {X}$ and $\overline {Y}$ . Then, applying Dedekind’s modular law, it follows that the normal subgroups $XG'$ and $YG'$ are properly contained in G; thus, X and Y cannot be contranormal in G. Therefore, X and Y are both normal-by-finite in G, and hence, $B=XY$ is likewise a normal-by-finite subgroup of G, which is a contradiction. Thus, $\overline {B}$ cannot be generated by two proper subgroups, so it is either a group of type $p^\infty $ or a cyclic p-group for some prime p.

If $\overline {B}$ is a group of type $p^\infty $ , then also $G/G'$ is a group of type $p^\infty $ . However, since $G'$ is a proper normal subgroup of G, all its subgroups are normal-by-finite in G. Moreover, if $\langle x \rangle $ is a cyclic subgroup of G which is not contained in $G'$ , it follows that $\langle x \rangle G'$ is a proper normal subgroup of G and, hence, all its subgroups are normal-by-finite in G. In particular, $\langle x \rangle $ is normal-by-finite in G, so G is a group of type (ii). If $\overline {B}$ is a cyclic p-group, then also $G/G'$ is a cyclic p-group and G is a group of type (iii).

Conversely, suppose that G satisfies one of the conditions (i)–(iii). Clearly, if condition (i) holds, every subgroup of G is either normal-by-finite or contranormal. Assume now that G satisfies condition (ii) and let H be a subgroup of G. Since every perfect quotient of G is trivial, if $G=G'H$ , then H is contranormal in G. However, if $G \neq G'H$ , it follows that $G'H/G'$ is a proper subgroup of $G/G'$ and, hence, $G'H/G'$ is a cyclic group. Therefore, $H/(H \cap G')$ is likewise cyclic and $H= (H \cap G')\langle y \rangle $ for some $y \in G$ . As all subgroups of $G'$ and all cyclic subgroups of G are normal-by-finite in G, also H is normal-by-finite in G, so that G has the required property. Finally, assume that G satisfies condition (iii) and let H be a subgroup of G. Since every perfect quotient of G is trivial, if $G=G'H$ , then H is contranormal in G. However, if $G \neq G'H$ , then $G'H$ is contained in $G' \langle x^p \rangle $ , so that H is normal-by-finite in G and the statement is proved.

Proof. Clearly, as N is a proper normal subgroup of G, any subgroup of infinite rank of N cannot be contranormal in G and, hence, is normal-by-finite in G. Therefore, [Reference De Falco, de Giovanni and Musella5, Theorem C] ensures that every subgroup of N is normal-by-finite and N is abelian-by-finite. In particular, N contains an abelian characteristic subgroup of finite index, so without loss of generality we may assume that N is abelian. Let X be any subgroup of finite rank of N. Since N has infinite rank, it contains subgroups U and V of infinite rank such that

$$ \begin{align*} \langle U,V \rangle = U \times V \end{align*} $$

and

$$ \begin{align*} X \cap \langle U,V \rangle = \{1\}. \end{align*} $$

As the indices $|U : U_G| $ and $|V : V_G|$ are finite, it follows that $U_G$ and $V_G$ have infinite rank. Therefore, the products $XU_G$ and $XV_G$ are normal-by-finite in G, and hence, also

$$ \begin{align*} X= XU_G \cap XV_G \end{align*} $$

is finite over its core $X_G$ .

Proof. Let X be any noncontranormal subgroup of finite rank of G. If $X^G$ has infinite rank, by Lemma 3.3, every subgroup of $X^G$ is normal-by-finite in G and so is X. Assume now that $X^G$ has finite rank. Let N be a proper normal subgroup of infinite rank of G; then N contains an abelian subgroup A of infinite rank (see [Reference De Falco, de Giovanni and Musella4, Lemma 5]). Thus, there exists a subgroup B of A such that B has infinite rank and $X^GB$ is a proper subgroup of G (see [Reference De Luca and Di Grazia13, Lemma 2.3]). Clearly, B cannot be contranormal in G and, hence, it is normal-by-finite in G. It follows that the product $X^GB$ is finite over its core $(X^GB)_G$ , so that $(X^GB)_G$ has infinite rank and Lemma 3.3 ensures that all its subgroups are normal-by-finite in G. In particular, X is normal-by-finite in G and the statement is proved.

Proof. We may assume that the locally soluble group G has infinite rank, so that it contains a proper normal subgroup of infinite rank. Therefore, it follows from Lemma 3.4 that every subgroup of G is either normal-by-finite or contranormal.

Proof. Let H be any large subgroup of N. Clearly, as N is a proper normal subgroup of G, H cannot be contranormal in G. Thus, every large subgroup of N is normal-by-finite in G and the statement follows from Lemma 2.1.

Proof. Let A be a large abelian subgroup of G and assume first that A is not contranormal in G. Therefore, A is normal-by-finite in G and, hence, its core $A_G$ is a large abelian normal subgroup of G. Suppose now that $A^G=G$ . Since large subgroups of $G'$ cannot be contranormal in G, if $A \cap G'$ has cardinality $\aleph $ , then it is normal-by-finite in G. Thus, the core $(A \cap G')_G$ is a large abelian normal subgroup of G. However, if $A \cap G'$ has cardinality strictly smaller than $\aleph $ , there exists a proper large subgroup B of A such that $A \cap G'$ is properly contained in B. Suppose that B is contranormal in G. It follows that $G=G'B$ and, applying Dedekind’s modular law,

$$ \begin{align*} B= (A\cap G')B=A\cap G'B=A, \end{align*} $$

which is a contradiction. Therefore, the large subgroup B is normal-by-finite in G and its core $B_G$ is a large abelian normal subgroup of G. This argument shows that, without loss of generality, we can assume that the large abelian subgroup A is normal in G.

Let H be any small noncontranormal subgroup of G. If $A \cap H^G$ has cardinality $\aleph $ , then $H^G$ is a proper large normal subgroup of G and, hence, Lemma 3.6 ensures that H is normal-by-finite in G. Thus, we may assume that $A\cap H^G$ has cardinality strictly smaller than $\aleph $ , so there exists a proper large subgroup C of A such that $A\cap H^G$ is properly contained in C. In particular, C is a large normal-by-finite subgroup of G. Suppose that $H^GC=G$ ; applying Dedekind’s modular law,

$$ \begin{align*} C= (A\cap H^G)C=A\cap H^GC=A, \end{align*} $$

which is a contradiction. It follows that the product $H^GC$ is a proper large normal-by-finite subgroup of G and, hence, its core $(H^GC)_G$ is a proper large normal subgroup of G. As the index $|C : A \cap (H^GC)_G|$ is finite, the intersection $A \cap (H^GC)_G$ is a large abelian subgroup of $(H^GC)_G$ and, by Lemma 3.6, every subgroup of $(H^GC)_G$ is normal-by-finite in G. In particular, H is normal-by-finite in G and the statement is proved.